Question

In the figure, an external force \(\vec{F}\) is holding a bob of mass \(500 . \mathrm{g}\) in a station ary position. The angle that the massless rope makes with the vertical is \(\theta=30.0^{\circ}\) a) What is the magnitude, \(F\), of the force needed to maintain equilibrium? b) What is the tension in the rope?

Short answer

Answer: The magnitude of the external force F required to maintain equilibrium is 2.83 N, and the tension in the rope is 5.66 N.

Step-by-step solution

Identify the forces acting on the bob

There are three forces acting on the bob: 1. The gravitational force, which is given by \(\vec{F}_{g} = m\vec{g}\), where m is the mass of the bob, and \(\vec{g}\) is the acceleration due to gravity. 2. The tension in the rope, represented by \(\vec{T}\), acts in the direction of the rope. 3. The external force, \(\vec{F}\), that is holding the bob in a stationary position.

Calculate the gravitational force

Since the mass of the bob is given as 500 g, we need to convert it to kg: \(m = 500\:g = 0.5\: kg\) Now, calculate the gravitational force acting on the bob using the standard value of gravitational acceleration, \(g \approx 9.81\: m/s^2\): \(\vec{F}_{g} = m\vec{g} = 0.5\: kg \times 9.81\: m/s^2 = 4.905\: N\)

Break down the tension force into its components

We need to break down the tension force into its vertical and horizontal components. \(\vec{T_v} = T\cos{\theta}\) (vertical component) \(\vec{T_h} = T\sin{\theta}\) (horizontal component)

Apply equilibrium conditions

Since the bob is in equilibrium, the net force in both horizontal and vertical directions should be zero. For the vertical direction: \(\vec{T_v} - \vec{F}_{g} = 0\) \(T\cos{30^{\circ}} - 4.905\: N = 0\) For the horizontal direction: \(\vec{T_h} - \vec{F} = 0\) \(T\sin{30^{\circ}} - F = 0\)

Solve for the external force F (part a)

First, solve the horizontal equilibrium equation for F: \(F = T\sin{30^{\circ}}\) Since \(\sin{30^{\circ}} = 0.5\), we can rewrite the equation as: \(F = 0.5T\)

Solve for the tension in the rope T (part b)

Now, use the vertical equilibrium equation to solve for tension T: \(T\cos{30^{\circ}} - 4.905\: N = 0\) Since \(\cos{30^{\circ}} = \frac{\sqrt{3}}{2}\), we can rewrite the equation as: \(T\frac{\sqrt{3}}{2} = 4.905\: N\) Now, solve for T: \(T = \frac{4.905\: N}{\frac{\sqrt{3}}{2}}\) \(T = 5.66\: N\) Finally, substitute the value of T back into the equation from Step 5 to find the magnitude of the external force F: \(F = 0.5(5.66\: N) = 2.83\: N\) The magnitude of the external force F required to maintain equilibrium is 2.83 N, and the tension in the rope is 5.66 N.